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Forces B/w Dislocations Consider two parallel (//) edge dislocations lying in the same slip plane. The two dislocations can be of same sign or different signs (a) Same Sign (on same slip plane)

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When the two dislocations are separated by large distance: The total elastic energy per unit length of the dislocation is given by: When dislocations are very close together: The arrangement can be considered approximately a single dislocation with Burgers vector = 2b In order to reduce the total elastic energy, same sign dislocations will repel each other (i.e., prefer large distance separation). (14.34) (14.35)

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Dislocations of opposite sign (on same slip plane) If the dislocations are separated by large distance: If dislocations are close together: Burgers vector = b - b = 0 Hence, in order to reduce their total energy, dislocations of opposite signs will prefer to come together and annihilate (cancel) each other.

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The same conclusions are obtained for dislocation of mixed orientations (a) and (b) above can be summarized as: –Like dislocations repel and –unlike dislocations attract

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Dislocations Not on the Same Slip Plane Consider two dislocations lying parallel to the z (x 3 ) -axis: In order to solve this: (a) We assume that dislocation “I” is at the origin (b) We then find the interaction force on dislocation “II” due to dislocation “I” x2x2 x1x1 x3x3 I II

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Recall Eqn. 14.29 Note that the dislocation at the origin (dislocation I) provides the stress field, while the Burgers vector and the dislocation length belongs to dislocation II Since is edge: Also b II is parallel to x 1 : Therefore, This means that b 2 = b 3 = 0 and b 1 = b 14.29

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Since t II is parallel to x 3, then This means that t 1 = t 2 = 0 and t 3 = 1 From Eqn. 14.31, we can write: Therefore

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But Therefore, F along the x 1 Direction is given as: 21 b This component of force is responsible for dislocation glide motion - i.e., for dislocation II to move along x 1 axis. 14.30

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F along the x 2 Direction is given as: - 11 b This component of force is responsible for climb (along x 2 ). At ambient (low) temperature, F x2 is not important (because, no climb). For edge dislocation, movement is by slip & slip occurs only in the plane contained by the dislocation line & its Burgers vector. 14.31

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Consider only component F x1 For x 1 >0: F x1 is negative (attractive) when x 1

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Hence Stable positions for two edge dislocations. 90 0 45 0

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Equations 14-30 and 14-31 can also be obtained by considering both the radial and tangential components. The force per unit length is given by: Because edge dislocations are mainly confined to the plane, the force component along the the x direction, which is the slip direction, is of most interest, and is given by: 14.32 14.33

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Eqn. 14-34 is same as 14-30. Figure 14-5 is a plot of the variation of F x with distance x, using equation 14-34. Where x is expressed in units of y. Curve A is for dislocations of the same sign; curve B is for dislocations of opposite sign. Note that dislocations of the same sign repel each other when x > y, and attract each other when x < y. 14.34

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Figure 14-5. Graphical representation of Eq. (14-21). Solid curve A is for two edge dislocations of same sign. Dashed curve B is for two unlike two dislocations.

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Example: A dislocations lies parallel to [100] with Burgers vector b. Compute the force acting on the dislocation due to the stress field of a neighboring screw dislocation lying parallel to [001]. Assume that for the screw dislocations x2x2 x1x1 x3x3 Solution: S

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Let the screw dislocation be dislocation I at the origin. The stress field for screw dislocation is given by: based on the assumption, we have

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For the other dislocation

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(continued)

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(b) (a) Figure 14-6. (a) Diffusion of vacancy to edge dislocation; (b) dislocation climbs up one lattice spacing

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